Hans-Georg said:
Kennedy,
thanks for the excellent explanation of the slanted razor edge
test.
Certainly an ideal scanner would yield white pixels on one side,
black pixels on the other, and one grey pixel in each row.
Sorry about the delay getting back to you Hans-Georg, but I had other
priorities over the weekend.
An imaginary scanner with 100% MTF at the Nyquist of the sampling
density would have only one grey pixel in the transition from white to
black. However that is unlikely to be an ideal or optimal scanner
because such a device would also have a significant, if not 100%, MTF
above Nyquist. An ideal scanner is not the imaginary scanner - it will
always produce *more* than 1 grey pixel in the transition.
To see why this is the case, you just need to consider the effect on the
MTF of the main components, the CCD and the optic.
In the case of the optic, the MTF is limited both by diffraction and
design and manufacturing aberrations. If the lens is very good it will
approach the situation where the MTF is dominated by the diffraction
term only, and hence is called diffraction limited. Diffraction is a
function of the lens aperture and the wavelength of the light
transmitted by it. If the lens has a variable aperture then it may
become diffraction limited above a certain f/#, since restricting the
aperture will increase the amount of diffraction produced and also
reduces the aberrations from the outer parts of the lens. So the
diffraction limited lens is the best that you can get. Roughly
speaking, diffraction spreads each point in the image into a gaussian
profile with a diameter at 1/e from the peak of approximately 2.44 x f/#
x wavelength. So, if we have an f/4 optic then the diameter of the
diffraction disc at the 1/e amplitude is about 10 wavelengths - or 5um
in the green and 7um in the red. The significant point is that the MTF
of a real lens, even an ideal lens, will always smear the image
slightly.
In spatial frequency terms, diffraction produces an MTF which falls
almost linearly from 1 at 0cy/mm spatial frequency down to 0.1 at 70% of
the cut-off frequency then tapering down to zero at the cut-off
frequency itself of about 1/(wavelength x f/#). The point here is that
the MTF at any particular sampling density will always be less than
unity, and can actually be quite low from even an ideal lens, and still
produce useful information.
For the CCD, the main limitation is the physical size of the CCD cell
itself. For most CCDs the optical collection area of the cell is close
to 80% of the pitch unless specific steps have been taken to reduce it
by masking etc. or increase it by microlenses. So the typical CCD
cannot discern anything smaller than 80% of a pixel pitch - whether it
is an infinitely fine spot or a blur it makes no difference to the CCD,
which is much the same as blurring the image itself, since the signal
produced by the cell only represents the average signal across its area.
Now, you can see that if you ignore the lens completely - or have a lens
which has almost 100% MTF at the limiting resolution of the scanner,
then the finite CCD element size itself will produce one grey pixel
between the black and white transition, because the chances of you
aligning your edge exactly with the transition between one CCD cell and
the next is negligible. You will almost always get one cell which sees
part of the black and part of the white, resulting in a grey pixel.
Similarly, if you ignore the finite width of the CCD cells, and assume
that they are infinitely thin, then the limited resolution of the lens,
even an ideal diffraction limited lens, will result in the edge becoming
blurred and hence at least one pixel (and possibly more) being grey.
Since both of these effects are unavoidable in even a perfect scanner,
it is fairly obvious that an ideal scanner must produce more than one
grey pixel between black and white transitions. More on this later
though.
Since scanners cannot be ideal, we have a very simple quality
signal here. We can check how many grey pixels we see when we
enlarge an area around the razor edge.
You can check the number of grey pixels but it does not tell you very
much on its own. The number of grey pixels across an edge may well be a
measure of quality, but it is not a measure of resolution and the
"optimum" is neither the maximum nor the minimum. In order to interpret
such a measure, you need to understand what the optimum is. You have
made an assumption that it is "1", but without any reference or
explanation of what artefacts you are prepared to tolerate in order to
achieve that figure, which I suggest you would find more undesirable
than the slightly softer result of a higher number. The optimum
scanning edge transition is not the same as the optimum downsampling of
a transition scanned at a much higher resolution since the latter can
invoke optimum digital filters which are impossible to manufacture
optically in the real world.
How many grey pixels per line would be acceptable before one
would say that the resolution is unnecessarily high in relation
to the poor optics?
That isn't a very useful question, in that it is a question which does
not have a unique answer. For example, your original assumption is that
an ideal scanner would have one grey pixel between black and white.
However, as explained, this would require a 100% MTF right up to, and
beyond, Nyquist, with all of the artefacts that such heavy undersampling
would produce. If your scanner were a simple analogue system, such as a
film camera or just the lens, then you would assess its resolution using
conventional Rayleigh criteria, which is roughly the spatial frequency
at which the MTF has fallen to around 10% or so. If this was a
diffraction limited lens then the MTF would reduce to this level almost
linearly, as mentioned above. So, if you had a scanner where the MTF
reduced linearly to around 10% at the Nyquist point, then you would get
scanned images which looked pretty close to ideal optics with minimum
sampling artefacts. How many pixels would appear grey in the transition
between black and white for such a scanner? 3-4, depending on the
position of the edge relative to the sample points - substantially more
than your nominally ideal, yet this would be considered ideal in an
analogue system, such as the camera lens that made your original image.
However, it doesn't end there. The lens/CCD combination probably
produces an MTF which does not reduce linearly, but is a concave down
curve, falling more rapidly than linear around 1/3rd of the spatial
frequency and then flattening off at higher frequencies. Such a scanner
will still "resolve" the full Nyquist of the sampling density but will
produce a transition which may be many times that "optimum". So you
really can't say that just because you have a grey transition of 10
pixels that the scanner resolution is unnecessarily high in relation to
the optics - what matters is the shape of the transition, not the number
of pixels it extends across. Since the shape is actually quite
difficult to assess visually, the best solution is to fourier transform
the transition to produce the MTF directly. Then you can simply set a
threshold for the ultimate contrast you want in your final scan and
simply read off the spatial frequency that corresponds to on the MTF
curve as your useful resolution directly. However, you need to be
careful not to be too aggressive in your requirements for MTF, since
data below this threshold can still be enhanced using USM, particularly
if scanning with high bit depth, to produce an end MTF which exceeds the
acceptable threshold.
By comparison, imagine you had a very high resolution scanner - let's
choose something impractically large for example, say 100,000ppi - and
produced your 4000ppi output by 25x downsampling the original data from
that device. In this case your "ideal" may well be to have only one
pixel grey in the transition between a black and white edge. Why does
this approach have a different optimum from scanning directly at
4000ppi? The answer lies in what is practical in terms of the system
MTFs. Whilst it is not practical to realise wither an optic or a CCD
with an MTF of 100% up to Nyquist and 0% beyond it, it is certainly
possible to closely approximate such an MTF in a digital filter and
incorporate such a filter in the downsampling algorithm. So, just
because you can achieve such a transition on downsampled data, it does
not mean that it is achievable, nor indeed desirable, at the original
scan resolution.
